Difference between revisions of "2015 UNCO Math Contest II Problems/Problem 10"
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(a) <math>14</math> (b) <math>132</math> | (a) <math>14</math> (b) <math>132</math> | ||
− | (c) <math>\frac{1}{n + 1} \binom{2n}{n}</math> <math>= \binom{2n}{n}- \binom{2n}{n-1}= \binom{2n}{n}- \binom{2n}{n+1}=\frac{1}{2n+1} \binom{2n+1}{n}=\frac{(2n)!}{(n+1)!n!}</math> | + | (c) <math>\frac{1}{n + 1} \binom{2n}{n}</math> <math>= \binom{2n}{n}- \binom{2n}{n-1}= \binom{2n}{n}- \binom{2n}{n+1}=\frac{1}{2n+1} \binom{2n+1}{n}=\frac{(2n)!}{(n+1)!n!}</math> <math>=</math> nth [[Catalan number]] |
== See also == | == See also == |
Revision as of 01:10, 8 January 2023
Problem
(a) You want to arrange biologists of different heights in two rows for a photograph. Each row must have biologists. Height must increase from left to right in each row. Each person in back must be taller than the person directly in front of him. How many different arrangements are possible?
(b) You arrange biologists of different heights in two rows of , with the same conditions on height as in part (a). How many different arrangements are possible? Remember to justify your answers.
(c) You arrange biologists of different heights in two rows of , with the same conditions on height as in part (a). Give a formula in terms of for the number of possible arrangements.
Solution
(a) (b)
(c) nth Catalan number
See also
2015 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by BONUS | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
All UNCO Math Contest Problems and Solutions |